Lemma 42.62.4. Let $k$ be a field. Let $f : X \to Y$ be a proper morphism of schemes smooth over $k$. Then the gysin map exists for $f$ and $f_*(\alpha \cdot f^!\beta ) = f_*\alpha \cdot \beta$.

Proof. Observe that $X \to X \times _ k Y$ is an immersion of $X$ into a scheme smooth over $Y$. Hence the gysin map exists for $f$ (Definition 42.59.4). To prove the formula we may decompose $X$ and $Y$ into their connected components, hence we may assume $X$ is smooth over $k$ and equidimensional of dimension $d$ and $Y$ is smooth over $k$ and equidimensional of dimension $e$. Observe that $f^![Y]_ e = [X]_ d$ (see for example Lemma 42.59.8). Write $\alpha = c \cap [X]_ d$ and $\beta = c' \cap [Y]_ e$, see Lemma 42.62.2. We have

\begin{align*} f_*(\alpha \cdot f^!\beta ) & = f_*(c \cap f^!(c' \cap [Y]_ e)) \\ & = f_*(c \cap c' \cap f^![Y]_ e) \\ & = f_*(c \cap c' \cap [X]_ d) \\ & = f_*(c' \cap c \cap [X]_ d) \\ & = c' \cap f_*(c \cap [X]_ d) \\ & = \beta \cdot f_*(\alpha ) \end{align*}

The first equality by the result of Lemma 42.62.2 for $X$. By Lemma 42.59.7 we know that $f^!$ commutes with $c'$. The commutativity of the intersection product justifies switching the order of capping $[X]_ d$ with $c$ and $c'$ (via the lemma). Commuting $c'$ with $f_*$ is allowed as $c'$ is a bivariant class. The final equality is again the lemma. $\square$

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